# How do you solve  3(4 - x)(2x + 1)<0?

Mar 14, 2016

x > 4 and x < $- \frac{1}{2}$. In other words, x is outside $\left[- \frac{1}{2} , 4\right]$

#### Explanation:

Drop the positive factor 3.
$- 2 {x}^{2} + 7 x + 4 < 0$.
${\left(x - \frac{7}{4}\right)}^{2} > \frac{81}{16}$
$| x - \frac{7}{4} | > \frac{9}{4}$.
If $| x - a | > b , x > a + b \mathmr{and} x < a - b$
Here, $x > 4 \mathmr{and} x < - \frac{1}{2}$.